Lesson 5 of 8 / Practical and data skills
Describe both the centre and the spread
Can two samples have the same mean but tell different stories?
In this lesson: Calculate a mean and sample standard deviation, and explain what each describes.
About 6 min
The key ideaThe mean describes the centre. Standard deviation describes the spread of observations around that mean, in the same units as the measurements.
Work with the evidence
Same mean, different spread
Values: 3, 4, 5, 6, 7 cm. Mean = 5 cm in both samples. Sample standard deviation = 1.58 cm. Wider spacing increases spread without moving the mean.
Explanation
Calculate a sample mean by adding all values and dividing by the number of observations. A mean is a summary, not a replacement for the raw data. Two samples can have the same mean while one contains much more widely spread observations.
For the sample standard deviation supplied in the syllabus, subtract the mean from each observation, square each difference, add the squared differences, divide by n - 1, then take the square root. Squaring prevents positive and negative deviations from cancelling. Taking the square root returns the result to the measurement unit.
For 3, 4, 5, 6 and 7 cm, the mean is 5 cm. The squared deviations are 4, 1, 0, 1 and 4 cm2, summing to 10 cm2. Divide by 4 and take the square root: s is about 1.58 cm. These are sample observations, so follow the supplied sample formula.
A larger standard deviation means greater spread, not a larger mean, greater accuracy or proof of an experimental mistake. Spread may arise from biological differences and measurement variation. An instrument with a consistent offset can give tightly grouped but inaccurate readings.
A normal distribution is symmetric and bell-shaped, with many observations near its centre and fewer far from it. Some statistical methods rely on an approximately normal underlying distribution. A small sample does not establish that assumption merely because its mean and standard deviation can be calculated. Error bars labelled as standard deviations describe spread; overlap alone is not a substitute for the specified significance test.
Step by step
- 1
Find the mean
Add the observations and divide by n.
- 2
Measure squared deviations
Subtract the mean, square each difference and add them.
- 3
Complete and interpret
Divide by n - 1, take the square root and describe the spread in the original units.
Worked example
Work through the evidence
Compare samples 3, 4, 5, 6, 7 cm and 1, 3, 5, 7, 9 cm.
One way to explain it
Both means are 5 cm. The first sample has s = square root of (10/4), about 1.58 cm. The second has s = square root of (40/4), about 3.16 cm. The second sample has greater spread despite an identical mean.
Why this answer works
- State the shared centre.
- Calculate spread from deviations, not the range alone.
- Interpret the larger standard deviation without calling the data wrong.
Is this true? "A small standard deviation proves the measurements are accurate."
Values can cluster tightly around a biased value. Standard deviation describes spread, while accuracy concerns closeness to the true value.